# LC Circuit inductance problem

1. Dec 13, 2009

### 1st1

1. The problem statement, all variables and given/known data
A circuit consists of a capacitor and an inductor. The resistance in the circuit is small and can be neglected. Initially, at t = 0 s, the voltage across the capacitor is at its maximum of VC = 10 V, the charge stored in the capacitor is 1 mC. It is observed that the capacitor discharges to QC = 0 C after t = 0.25 ms. What is the inductance L of the inductor in the circuit?

http://online.physics.uiuc.edu/cgi/courses/shell/common/showme.pl?courses/phys212/oldexams/exam3/fa09/fig8.gif [Broken]

2. Relevant equations

Conservation of energy:

Energy Capacitor + Energy Inductor = Qmax^2 / 2C

3. The attempt at a solution

Initial Energy capacitor = .5*(Qmax^2/C) = .005
Final Energy Capacitor = 0

Initial Energy Inductor = 0
Final Energy Inductor = .5*L*I^2

So:

.005 = .5*L*I^2 where I = dQ/dt which is (1E-3)/(.25E-3) = 4

L = 6.25E-4 H which is wrong.

What am I doing wrong?
Thanks for any help!

Last edited by a moderator: May 4, 2017
2. Dec 13, 2009

### AEM

The inductor-capacitor circuit that you describe is analogous to a simple frictionless mass-spring system. It may be described by the following differential equation:

$$L \frac{d^2q}{dt^2} + \frac{1}{C}q = 0$$

The solution is $q = q_m cos(\omega t + \phi)$ which gives the charge on the capacitor at any time t.

For the initial condition you describe, the constant $\phi$ is zero and

$\omega = \sqrt{\frac{1}{LC}}$.

Now, the capacitor discharges in .25 msec. What is the value of $\omega t$ when q = 0? (Hint: it has to make the cosine = 0.) That is sufficient for you to calculate L.

Your mistake was assuming that the current is simply q/t.

Last edited by a moderator: May 4, 2017
3. Dec 13, 2009

### 1st1

Thank you thats perfect!